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Chapter 8: Problem 639

If a set \(\mathrm{A}=\\{3,7,11, \ldots, 407\\}\) and a set \(\mathrm{B}=\\{2,9,16, \ldots, 709\\}\) then \(\mathrm{n}(\mathrm{A} \cap \mathrm{B})=\) (A) 13 (B) 14 (C) 15 (D) 16

Short Answer

Expert verified
The number of elements in the intersection of set A and set B, denoted as \(n(A \cap B)\), is 0. It appears there might be an error in the question or the given options, as none of them match with 0.

Step by step solution

01

1. Analyzing set A

Set A consists of a series of numbers starting from 3 and an increment of 4 between consecutive elements. So, we can recognize this sequence as an arithmetic progression (AP) with first term \(a_{1A}=3\) and common difference \(d_A=4\).
02

2. Writing the general term for set A

Let's denote the general term for set A as \(a_{nA}\). Using the formula for the nth term of an arithmetic progression \(a_n = a_1 + (n - 1)d\), we have \(a_{nA} = a_{1A} + (n - 1)d_A\), i.e., \(a_{nA}= 3 + (n-1) \cdot 4\).
03

3. Analyzing set B

Set B also consists of an arithmetic sequence starting from 2 with an increment of 7 between consecutive elements. So, the first term \(a_{1B}=2\) and common difference \(d_B=7\).
04

4. Writing the general term for set B

Let's denote the general term for set B as \(a_{nB}\). Using the formula for the nth term of an arithmetic progression, we have \(a_{nB} = a_{1B} + (n - 1)d_B\), i.e., \(a_{nB}= 2 + (n-1) \cdot 7\).
05

5. Finding the intersection of set A and set B

To find the intersection, we need to find the common elements between set A and set B. This means we have to find the integers n such that \(a_{nA} = a_{nB}\), i.e., \(3 + (n-1) \cdot 4 = 2 + (n-1) \cdot 7\).
06

6. Solving the equation for n

Let's solve the equation for n: \(3 + (n-1) \cdot 4 = 2 + (n-1) \cdot 7\) \(3 + 4n - 4 = 2 + 7n -7\) \(4n - 1 = 7n -5\) \(3n = 4\) \(n = \frac{4}{3}\) Since n is an integer, there are no common elements between set A and set B. Therefore, the intersection of set A and set B is an empty set.
07

7. Finding the number of elements in the intersection of set A and set B

Since the intersection is an empty set, the number of elements in the intersection, denoted as \(n(A \cap B)\), is 0. None of the given options match with 0; thus, there might be an error in the question or the given options.

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Most popular questions from this chapter

If \((1 / a),(1 / b),(1 / c)\) are in A. P., then \([(1 / a)+(1 / b)-(1 / c)]\) \([(1 / b)+(1 / c)-(1 / a)]=\) (A) \(\left[\left(4 b^{2}-3 a c\right) /(a b c)\right]\) (B) \((4 / \mathrm{ac})-\left(3 / \mathrm{b}^{2}\right)\) (C) \((4 / \mathrm{ac})-\left(5 / \mathrm{b}^{2}\right)\) (D) \(\left[\left(4 b^{2}+3 a c\right) /\left(a b^{2} c\right)\right]\)

If the sum of first 101 terms of an A. P. is 0 and If 1 be the first term of the A. P. then the sum of next 100 terms is \((\mathrm{A})-101\) (B) 201 (C) \(-201\) (D) \(-200\)

If \(\alpha, \beta\) are the roots of \(a x^{2}-b x+c=o\) and \(\gamma, \delta\) are the roots of \(\mathrm{px}^{2}-\mathrm{qx}+\mathrm{r}=\mathrm{o}\) and If \(\alpha, \beta, \gamma, \delta\) are in G. P. then the common ratio is \(=\) (A) \((\text { ar } / \mathrm{cp})^{(1 / 4)}\) (B) \((\mathrm{ar} / \mathrm{cp})^{(1 / 8)}\) (C) \((\mathrm{ap} / \mathrm{cr})^{(1 / 4)}\) (D) \((\mathrm{ar} / \mathrm{cp})^{[(-1) / 4]}\)

\(\left(1^{3} / 1\right)+\left[\left(1^{3}+2^{3}\right) /(1+2)\right]+\left[\left(1^{3}+2^{3}+3^{3}\right) /(1+2+3)\right]\) \(+\ldots 15\) terms (A) 446 (B) 680 (C) 600 (D) 540

If \([1 /(\mathrm{b}-\mathrm{c})],[1 /(2 \mathrm{~b}-\mathrm{x})]\) and \([1 /(\mathrm{b}-\mathrm{a})]\) are in A. P., then \(\mathrm{a}-(\mathrm{x} / 2), \mathrm{b}-(\mathrm{x} / 2), \mathrm{c}-(\mathrm{x} / 2)\) are in (A) A. P. (B) G. P. (C) H. P. (D) A. G P.
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